One base of a trapezoid is decreasing at a rate of $8$ kilometers per second and the height of the trapezoid is increasing at a rate of $5$ kilometers per second. The other base of the trapezoid is fixed at $4$ kilometers. At a certain instant, the decreasing base is $12$ kilometers and the height is $2$ kilometers. What is the rate of change of the area of the trapezoid at that instant (in square kilometers per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $32$ (Choice B) B $-22$ (Choice C) C $-32$ (Choice D) D $22$
Solution: Setting up the math Let... $b_1(t)$ denote the decreasing base of the trapezoid at time $t$, $h(t)$ denote the height of the trapezoid at time $t$, and $A(t)$ denote the area of the trapezoid at time $t$. We are given that $b_1'(t)=-8$ and $h'(t)=5$ (notice that $b_1'$ is negative). We are also given that that $b_1(t_0)=12$ and $h(t_0)=2$ for a specific time $t_0$. We are also given that the other base $b_2$ is $4$ kilometers. We want to find $A'(t_0)$. Relating the measures The measures relate to each other through the formula for the area of a right trapezoid: $\begin{aligned} A(t)&=\dfrac{b_1(t)+b_2}{2}\cdot h(t) \\\\ &=\dfrac12b_1(t)h(t)+\dfrac{b_2}{2}h(t) \end{aligned}$ Plugging $b_2=4$, we get the following equation: $A(t)=\dfrac12b_1(t)h(t)+2h(t)$ We can differentiate both sides to find an expression for $A'(t)$ : $\dfrac12 b_1'(t)h(t)+\dfrac12b_1(t)h'(t)+2h'(t)$ Using the information to solve Let's plug ${b_1'(t_0)}={-8}$, ${h(t_0)}={2}$, ${b_1(t_0)}={12}$, and $C{h'(t_0)}=C{5}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=\dfrac12{b_1'(t_0)}{h(t_0)}+\dfrac12{b_1(t_0)}C{h'(t_0)}+2C{h'(t_0)} \\\\ &=\dfrac12 ({-8})({2})+\dfrac12({12})(C{5})+2(C{5}) \\\\ &=32 \end{aligned}$ In conclusion, the rate of change of the area of the trapezoid at that instant is $32$ square kilometers per second. Since the rate of change is positive, we know that the area is increasing.